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 November 1st, 2011, 12:41 PM #1 Newbie   Joined: Nov 2011 Posts: 4 Thanks: 0 Finding k so that a given quadratic has real roots. I need help with these questions. Any help would be greatly appreciated. Thanks. 1. Find the values of k such that the roots of x-k(x-1)(x-2)=0 are real roots. [color=#FFFFFF] 2. A mural is to be painted on a wall that is 15 meters long an 12 meters high. A border of uniform width is to surround the mural. If the mural is to cover 75% of the area of the wall, how wide must the border be, to the nearest hundredth of a meter? 3. Write the reciprocal of a+bi in the simplest form. 4. Write a quadratic equation whose roots are 3+2i/2 and 3-2i/2. 5. For the following relation, determine the value of y in terms of x. 4x^2 - 4xy + y^2 - 2x + y - 6 = 0[/color]
 November 1st, 2011, 12:44 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Help Please :) 1. Get this equation in standard form: $ax^2 + bx + c= 0$ a, b, and c will be in terms of integers and "k". The root(s) of this equation will be real when $b^2 - 4ac \geq 0$
November 1st, 2011, 12:51 PM   #3
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 Originally Posted by The Chaz 1. Get this equation in standard form: $ax^2 + bx + c= 0$ a, b, and c will be in terms of integers and "k". The root(s) of this equation will be real when $b^2 - 4ac \geq 0$
I'm left with this: x-k$(x^2 - 3x +2)$.

I don't know what to do with the x-k.

 November 1st, 2011, 12:54 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Help Please :) $x-k(x^2 - 3x +2)$ If you're confused, then I'm concerned! Multiply... $x-k(x^2 - 3x +2)= x -kx^2 +3kx - 2k = 0$ Group the x's together... $x-k(x^2 - 3x +2)= x -kx^2 +3kx - 2k = -kx^2 + (3k + 1)x -2k = 0$ Can you take it from here?
November 1st, 2011, 01:09 PM   #5
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 Originally Posted by The Chaz $x-k(x^2 - 3x +2)$ If you're confused, then I'm concerned! Multiply... $x-k(x^2 - 3x +2)= x -kx^2 +3kx - 2k = 0$ Group the x's together... $x-k(x^2 - 3x +2)= x -kx^2 +3kx - 2k = -kx^2 + (3k + 1)x -2k = 0$ Can you take it from here?
Thanks, I was confused for some reason.

Can you help me solve the others?

November 1st, 2011, 01:20 PM   #6
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 Originally Posted by tmac20 ... Thanks, I was confused for some reason. Can you help me solve the others?
Yes, but let's do it this way:
Create a separate thread for each, so we can give them each an appropriate title and organize the replies.

I'll lock this thread so you can get started on that.

 November 1st, 2011, 01:29 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Help Please :) 3.) The reciprocal or multiplicative inverse of a + bi is: $\frac{1}{a+bi}\cdot\frac{a-bi}{a-bi}=\frac{a-bi}{a^2+b^2}$ 4.) We may write: $$$x-\frac{3+2i}{2}$$$$x-\frac{3-2i}{2}$$=0$ $$$2x-(3+2i)$$$$2x-(3-2i)$$=0$ $4x^2-2(3-2i)x-2(3+2i)x+9+4=0$ $4x^2-12x+13=0$ 5.) We are given: $4x^2-4xy+y^2-2x+y-6=0$ Write as quadratic on y: $y^2+(1-4x)y+2$$2x^2-x-3$$=0$ $y=\frac{4x-1\pm\sqrt{(1-4x)^2-8$$2x^2-x-3$$}}{2}$ $y=\frac{4x-1\pm\sqrt{1-8x+16x^2-16x^2+8x+24}}{2}$ $y=\frac{4x-1\pm5}{2}$ $y=\frac{4x-6}{2}=2x-3$ $y=\frac{4x+4}{2}=2(x+1)$ edit: oops! I will move the responses to the appropriate threads once created!

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